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Coulombs Law-what now?

  1. Feb 4, 2010 #1
    Coulombs Law--what now?

    1. The problem statement, all variables and given/known data

    Three point charges are fixed in place in a right triangle. What is the electric force on the q = -0.64 µC charge due to the other two charges? (Let Q1 = +0.71 µC and Q2 = +1.1 µC.)


    magnitude N


    direction ° above the positive x-axis

    2. Relevant equations

    F=kq1q2 / r^2

    K=8.99 x 10^9

    3. The attempt at a solution

    The diagram from the book shows a right triangle where q is the 90 degrees vertex along X-axis. the hypotenuse = 10 cm; r (y)=8.0 cm; and between q & Q2 =6 cm.

    I have made free body diagrams and know: q = -0.64 µC; on the y axis Q1= +0.71 µC and Q2 = +1.1 µC on the X-axis

    plugging the info into the formulas I get:

    F (y)= (8.99x10^9)(0.71)(0.64) / 8^2 = 6382900

    F (x)= (8.99x10^9)(1.1)(0.64) / 6^2 = 175804444.4

    Don't these numbers seem rather big? Did I miss a conversion somewhere?

    I am also unsure of what to do next.

    Thanks in advance for your help.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 4, 2010 #2
    Re: Coulombs Law--what now?

    Those numbers are definitely plausible. You have to remember that k = 9e9. So if those numbers are correct, what you want to do is draw the vectors associated with them and use cosine law most likely to find the summation of those vectors.
  4. Feb 4, 2010 #3
    Re: Coulombs Law--what now?

    These numbers seem big, because you're not converting any of the units to the standard SI! Convert the microCoulombs to Coulombs and the centimeters to meters. Once you correct your order of magnitude by proper unit conversion, your next step should be trigonometric. Think of the x and y components as sides of a triangle with the hypotenuse equal to the magnitude.
    Last edited: Feb 4, 2010
  5. Feb 4, 2010 #4
    Re: Coulombs Law--what now?

    thanks I'll try it again.
  6. Feb 5, 2010 #5
    Re: Coulombs Law--what now?

    Hi. After converting as per your advice, I have the following numbers:

    F (Q1q) = (8.99 x 10^9)(7.1 x 10^-5)(6.4 x 10^-5) / .08^2 = 6382.9

    F (Q2q) = (8.99 x 10^9)(1.1 x 10^-4)(6.4 x 10^-5) / .06^2 = 17580.44444

    Please check this for accuracy (I'm new to the calculator apps--it's likely I made errors).

    If conversions are correct--what is my next step? Please give example. Thanks.
  7. Feb 5, 2010 #6
    Re: Coulombs Law--what now?

    Your units are still converted incorrectly.
    [tex]\frac{.71\mu C}{1}*\frac{1C}{10^6\mu C}=710x10^{-9}\neq 7.1x10^{-5} C[/tex]

    I believe your other charge conversions are faulty too.

    http://img682.imageshack.us/img682/6978/70405368.jpg [Broken]
    Assuming you did your math right and both your forces are positive, this is a picture of what you're trying to find in your next step. Instead of showing the answer as two components - one vector at 0 degrees and the other at 90 degrees - you do vector addition to represent your answer as one vector. This single vector means you need a magnitude and a direction.

    To find the magnitude, use 90 degree trigonometry. To find the angle, do the same. EDIT: I also just realized I typed the two components in reverse. The x component is supposed to be on the x-axis, and the y-component is supposed to be on the y-axis.
    Last edited by a moderator: May 4, 2017
  8. Feb 7, 2010 #7
    Re: Coulombs Law--what now?

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